My pointer p is inside a function, will i get memory leak with this code.
for(k=0;k< 3;k++)
{
int *p=NULL;
int val = bBreak[k+1] - bBreak[k];
p = new int [val+1];
p = &buff[bBreak[k]];
for(int i=0;i< val;i++)
{
cout<<"\n"<<p[i]<<endl;
}
}
Yes! You never free the memory. You should call delete/delete[] for every piece of memory you allocate with new/new[].
Yes, you will
p = new int [val+1]; //allocate array on the heap
p = &buff[bBreak[k]]; //new allocated array is leaked because you lost the pointer to it
//and you are not able to call 'delete[]' to free the memory
Generally, every call to operator new should be paired with call of operator delete or delete[]
Yes. You must delete every memory you allocate with new.
p = new int [val+1];
p = &buff[bBreak[k]]; // here you lose track of the memory you've just allocated
If you don't want to do memory management by-hand, use a std::vector<int>.
Related
Please don't come and say "jUst read the error", it doesn't specify it, that's why Im asking here.
int main()
{
int* p, * q;
int a = 10;
p = new int;
p = &a;
delete p;
}
p = new int
In this line, you dynamically allocate the memory that you need to delete.
p = &a
But in this line, you are not allocating new memory, you are pointing p at stack memory, and leaking the memory that you previously allocated.
What is the difference between these pointers?
I know that this one is going to be stored on the heap, even though a pointer is only 8 bytes anyways, so the memory is not important for me.
int* aa = new int;
aa = nullptr;
and this one is going to be stored on the stack.
int* bb = nullptr;
They both seem to work the same in my program. Is there any difference apart from memory allocation? I have a feeling that the second one is bad for some reason.
2) Another question which is somewhat related:
Does creating a pointer like that actually take more memory? If we take a look at the first snippet, it creates an int somewhere (4 bytes) and then creates a pointer to it (8 bytes), so is it 12 bytes in total? If yes are they both in the heap then? I can do this, so it means an int exists:
*aa = 20;
Pointers are integers that just indicate a memory position, and a type (so they can only point to variables of that type).
So in your examples, all pointers are stored in the stack (unless they are global variables, but that is another question). What they are pointing to is in the heap, as in the next example.
void foo()
{
int * ptr = new int(42);
// more things...
delete ptr;
}
You can have a pointer pointing into the stack, for example, this way:
void foo()
{
int x = 5;
int * ptr = &x;
// more things...
}
The '&' operator obtains the memory position of the variable x in the example above.
nullptr is the typed equivalent to old NULL. They are a way to initialize a pointer to a known and secure value, meaning that they are not pointing to anything else, and that you can compare whether they are NULL or not.
The program will accept pointers pointing to the stack or the heap: it does not matter.
void addFive(int * x)
{
*x += 5;
}
void foo()
{
int x = 5;
int * ptr1 = &x;
int * ptr2 = new int(42);
addFive( ptr1 );
addFive( ptr2 );
addFive( &x );
printf( "%d\n", *ptr1 );
printf( "%d\n", *ptr2 );
// more things...
delete ptr2;
}
The only difference is that the C runtime will keep structures telling how much memory has been spent in the heap, and therefore storing variables in the heap comes at a cost in performance. On the other hand, the stack is always limited to a fixed amount of memory (relatively small), while the heap is much larger, allowing you to store big arrays, for example.
You could take a look at C-Sim, which simulates memory in C (disclaimer: I wrote it).
Hope this helps.
I just learned pointer and delete pointer in class for C++. I tried this code by my own
# include<iostream>
using namespace std;
int main(){
int num = 10;
int *p = new int;
p = #
cout << *p << endl;
delete p;
cout << num << endl;
return 0;
}
After deleting the pointer p, I cannot print the value of num. But if I delete p at the very end of the program, cout << num << endl; will give me 10. Anyone knows where I did run?
You first leaked a pointer
int *p = new int;
p = # // You just leaked the above int
then illegally deleted something you did not new
delete p; // p points to num, which you did not new
You have already received a couple of good answers that point out the mistake, but I read a deeper misunderstanding of the allocation and deallocation of heap vs stack variables.
I realised this has become a pretty long post, so maybe if people think it is useful I should put it as a community Wiki somewhere. Hopefully it clarifies some of your confusion though.
Stack
The stack is a limited and fixed size storage. Local variables will be created here if you don't specify otherwise, and they will be automatically cleaned up when they are no longer needed. That means you don't have to explicitly allocate them - they will start existing the moment you declare them. Also you don't have to deallocate them - they will die when they fall out of scope, loosely speaking: when you reach the end brace of the block they are defined in.
int main() {
int a; // variable a is born here
a = 3;
a++;
} // a goes out of scope and is destroyed here
Pointers
A pointer is just a variable, but instead of an int which holds a whole number or a bool which holds a true/false value or a double which holds a floating point, a pointer holds a memory address. You can request the address of a stack variable using the address operator &:
{
int a = 3, b = 4;
int* p = &a; // p's value is the address of b, e.g. 0x89f2ec42
p = &b; // p now holds the address of b, e.g. 0x137f3ed0.
p++; // p now points one address space further, e.g. 0x137f3ed4
cout << p; // Prints 0x137f3ed4
} // Variables a, b and p go out of scope and die
Note that you should not assume that a and b are "next to" each other in memory, or that if p has a "used" address as its value then you can also read and write to the address at p + 1.
As you probably know, you can access the value at the address by using the pointer indirection operator, e.g.
int* p = &a; // Assume similar as above
*p = 8;
cout << a; // prints 8
cout << &a << p; // prints the address of a twice.
Note that even though I am using a pointer to point at another variable, I don't need to clean up anything: p is just another name for a, in a sense, and since both p and what it points to are cleaned up automatically there is nothing for me to do here.
Heap
The heap memory is a different kind of memory, which is in theory unlimited in size. You can create variables here, but you need to tell C++ explicitly that you want to do so. The way to do this is by calling the new operator, e.g. new int will create an integer on the heap and return the address. The only way you can do something sensible with the allocated memory, is save the address this gives you. The way you do this, is store it in a pointer:
int* heapPtr = new int;
and now you can use the pointer to access the memory:
*heapPtr = 3;
cout << heapPtr; // Will print the address of the allocated integer
cout << *heapPtr; // Will print the value at the address, i.e. 3
The thing is that variables created on the heap will keep on living, until you say you don't need them anymore. You do that by calling delete on the address you want to delete. E.g. if new gave you 0x12345678 that memory will be yours until you call delete 0x12345678. So before you exit your scope, you need to call
delete heapPtr;
and you will tell your system that the address 0x12345678 is available again for the next code that comes along and needs space on the heap.
Leaking memory
Now there is a danger here, and that is, that you may lose the handle. For example, consider the following:
void f() {
int* p = new int;
}
int main() {
f();
cout << "Uh oh...";
}
The function f creates a new integer on the heap. However, the pointer p in which you store the address is a local variable which is destroyed as soon as f exits. Once you are back in the main function, you suddenly have no idea anymore where the integer you allocated was living, so you have no way to call delete on it anymore. This means that - at least for the duration of your program - you will have memory that according to your operating system is occupied, so you cannot use it for anything else. If you do this too often, you may run out of memory even though you can't access any of it.
This is one of the errors you are making:
int* p = new int;
allocates a new integer on the heap and stores the address in p, but in the next line
p = #
you overwrite that with another address. At this point you lose track of the integer on the heap and you have created a memory leak.
Freeing memory
Aside from freeing memory not often enough (i.e. not instead of once), the other error you can make is freeing it too often. Or, to be more precise, you can make the error of accessing memory after you have told your OS you don't need it anymore. For example, consider the following:
int main() {
int* p = new int;
*p = 10;
delete p; // OK!
*p = 3; // Errr...
}
That last line is very wrong! You have just returned the memory allocated when you called delete, but the address is still stored in p. After you call delete, your OS is allowed to re-allocate the memory at any time - for example, immediately after another thread could call new double and get the same address. At that point, if you write *p = 3 you are therefore writing to memory that is no longer yours which may lead to disaster, if you happen to overwrite the location in memory where the nuke's launch codes are stored, or nothing may happen at all because the memory is never used for anything else before your program ends.
Always release your own memory, and nothing but your own memory
We have concluded the following: memory allocated on the stack is not yours to claim, and not yours to release. Memory allocated on the heap is yours to claim, but you must also release it once and only once.
The following examples are incorrect:
{
int a = 3;
int* p = &a;
delete a;
} // Uh oh... cannot clean up a because it is not ours anymore!
{
int* p = new int;
delete p;
*p = 3; // Uh oh, cannot touch this memory anymore!
delete p; // Uh oh, cannot touch this memory anymore!
}
Why does it print 10?
Well, to be honest, you were just "lucky" there. Actually, the way your operating system manages memory, is generally pretty lazy. When you tell it "I would like some memory" it doesn't zero it for you. That is why it is a bad idea to write
int main() {
int a;
a = a + 3;
cout << a;
}
You get allocated a variable a somewhere in the memory, but the value of a will be whatever was in that memory location. It might be zero, or some random number that depends on how the bits fell when you booted your computer. That is why you should always initialize the variable:
int a = 0;
Similarly, when you say "I don't need this memory" anymore, the OS doesn't zero it. That would be slow and unnecessary: all it needs to do is mark the memory as "free to be re-allocated". So if you give it back and access it immediately afterwards, the probability that it has not been re-allocated yet is pretty large. Therefore
int* p = new int;
*p = 10;
delete p;
cout << *p;
is not guaranteed to print 10. The address p is pointing to may have been (partially) taken (and initialized!) by someone else immediately after the delete. But if it hasn't, the memory will still contain the value 10 there so even though it isn't yours anymore, C++ will still allow you to access it. Basically, when you are using pointers, you are telling it "trust me, I'm a programmer - you don't need to do all kinds of slow checks to make sure I'm staying where I'm supposed to be, instead I'll be careful about that myself!"
using namespace std;
int main(){
int num = 10; // a) an int is created on stack
int *p = new int; // b) another int is allocated on heap
p = # // c) address of int from stack is assigned to p and the pointer
// allocated in b) is leaked: as nothing points to it anymore,
// it can't be deleted
cout << *p << endl;
delete p; // d) deleting a pointer that was not dynamically allocated
// and is pointing to stack.
cout << num << endl;
return 0;
}
First of all, i had some confusion on pointer deletion. When i delete pointer a, i don't understand why int x is also deleted.. because i already assign a = f(which is the pointer pointing to y)
int x = 5;
int y = 3;
int *a = &x;
int *f = &y;
a = f;
delete a;
cout << x; // Segmentation fault (core dumped)
Second, if i want to assign value to a class with a char * pointer. How can i do?
class Foo{
public:
char * name;
void setName(char * const newName){
name = newName;
}
};
void getInput()
{
char temp[5];
Foo s;
char * x = new char[5]; // four words
cin >> temp;
strcpy(x,temp);
s.setName(x); // should i delete x pointer
// afterward because this function
// is going to terminated ?
}
You can only delete something that has been allocated with new. In your case, you are trying to delete something else, so you get undefined behaviour.
If you want your Foo class to "take ownership" of the pointer and be responsible for deleting it, then you should delete name in the class destructor.
Although a is a pointer, you can't delete it. It points to a memory block on the stack (not heap), it cannot be delete-ed.
The 'delete' operator is used to free up user-allocated space on the heap when you no longer have use of it. All allocated memory will be delete-ed once your application has ended automatically.
But... say you have a service that runs in the background in an endless loop and you allocate a space on the heap without delete it -> this allocated memory will most likely not be delete-ed (depends on the OS and type of pointer - volatile or not). It is in the developer's responsibility to make sure all allocated memory is delete-ed once his/her object is no longer in use (end of block or in the destructor).
here if I use delete or delete[] the output is still 70. Can I know why?
#include<iostream>
using namespace std;
int main()
{
int* c = new int[100];
for(int i=0; i<98; i++)
{
c[i] = i;
}
cout<<c[70]<<endl;
delete[] c;
or
delete c;
cout<<c[70]<<endl; //outputs 70 even after delete[] or delete
return 0;
}
Accessing deleted memory is undefined behavior. Deleting with the wrong delete is also UB. Any further discussion is pointless in the sense that you cannot reliably expect any outcome.
In many cases, UB will just do the "correct" thing, but you need to be aware that this is completely "by chance" and could change with another compiler, another version of the same compiler, the weather... To get correct code, you need to avoid all cases of UB, even those that seemingly work.
Using new will just allocate some memory to your program and return a pointer pointing at the said memory address, reserving as much memory as needed for the datatype. When you use delete later, it "frees" the memory, but doesn't delete it's content. If you had an int with the value 70 stored at that address, it will still contain 70, until another application wants some memory, gets said address and puts another value in there.
If you use new to allocate memory for an array, you will reserve following blocks of memory until there are enough blocks for your specified array length.
Let's say you do the following:
int main() {
int* array = new int[10]; // array now points to the first block of the allocated memory
delete array; // since array points to the first block of the array, it will only free that block, but nothing else, causing a memory leak
delete[] array; // will free all memory allocated by the previous new
// Note that you should never free allocated memory twice, like in this code sample. Using delete on already freed memory is undefined behaviour!
]
Always use delete for single variables and delete[] for arrays.
A demonstration of your problem:
int main() {
int* c = new int[10]; // We allocate memory for an array of 10 ints
c[0] = 1; // We set the value of the first int inside the array to 1
delete[] c;
/*
* We free the previously allocated memory.
* Note that this does not delete the CONTENT of the memory!
* c does still point towards the first block of the array!
*/
std::cout << c[0];
/*
* Firstly, this is undefined behaviour (Accessing deallocated memory).
* However, this will output 1,
* unless some other process allocated the memory at the address
* and filled it with another value already. (Very unlikely)
*/
return 0;
}
If you want to delete / overwrite the content of the deleted memory, you can use std::memset.
Example:
#include <cstring>
int main() {
std::size_t length = 10;
int* c = new int[length];
c[0] = 1;
std::cout << c[0] << std::endl; // Will output 1
std::memset( c, 0, length ); // Fill the memory with 0 bytes
delete[] c; // Now we free the array's memory
std::cout << c[0] << std::endl; // Will output 0
}
As others pointed its undefined behaviour and anything can happen.
These can be easily caught with the help of tools like valgrind.